Q. how to calculate formal charge

Definition and formula. Formal charge is an electron bookkeeping tool that helps choose the best Lewis structure. Use the formula for each atom:

\[ \text{Formal charge} = V – N – \frac{B}{2} \]

Explanation of symbols. V is the number of valence electrons the neutral atom brings. N is the number of nonbonding electrons assigned to that atom in the Lewis structure. B is the total number of bonding electrons shared in bonds to that atom. Bonding electrons means all electrons in bonds that the atom participates in, counting each shared electron individually.

Step by step procedure:

Step 1. Count V. Determine the atom’s valence electrons from the periodic table. For main group elements, this is the group number.

Step 2. Count N. Count the electrons in lone pairs on that atom. Each lone pair contributes two nonbonding electrons.

Step 3. Count B. Count all electrons in bonds attached to the atom. A single bond contributes two bonding electrons, a double bond contributes four bonding electrons, and a triple bond contributes six bonding electrons.

Step 4. Apply the formula. Compute \(\text{Formal charge} = V – N – \frac{B}{2}\). The result can be positive, negative, or zero. The sum of formal charges for all atoms equals the total charge of the molecule or ion.

Example 1 carbon dioxide carbon atom. Write the Lewis structure with carbon double bonded to two oxygens. For the carbon atom count V equals 4. Nonbonding electrons N equals 0 on carbon. Bonding electrons B equals 8 because there are two double bonds, each double bond equals four electrons, total eight. Compute the formal charge:

\[ \text{Formal charge}_{\text{C}} = 4 – 0 – \frac{8}{2} = 4 – 4 = 0 \]

Example 1 carbon dioxide oxygen atom. For an oxygen that is double bonded to carbon count V equals 6. Nonbonding electrons N equals 4 because there are two lone pairs on that oxygen. Bonding electrons B equals 4 from the double bond. Compute the formal charge:

\[ \text{Formal charge}_{\text{O}} = 6 – 4 – \frac{4}{2} = 6 – 4 – 2 = 0 \]

Conclusion for CO2. All atoms have formal charge zero, which matches the neutral molecule.

Example 2 nitrate ion NO3 minus. Use a common Lewis structure with one N double bonded to one O and single bonded to two O atoms. The overall charge is negative one. Compute for nitrogen. V equals 5 for nitrogen. N equals 0 nonbonding electrons on nitrogen in this structure. B equals 8 bonding electrons because nitrogen participates in one double bond worth four electrons and two single bonds worth two electrons each, total eight. Compute nitrogen formal charge:

\[ \text{Formal charge}_{\text{N}} = 5 – 0 – \frac{8}{2} = 5 – 4 = +1 \]

Compute for the double bonded oxygen. V equals 6. N equals 4 from two lone pairs. B equals 4 from the double bond. Compute its formal charge:

\[ \text{Formal charge}_{\text{O(double)}} = 6 – 4 – \frac{4}{2} = 6 – 4 – 2 = 0 \]

Compute for a single bonded oxygen. V equals 6. N equals 6 from three lone pairs. B equals 2 from the single bond. Compute its formal charge:

\[ \text{Formal charge}_{\text{O(single)}} = 6 – 6 – \frac{2}{2} = 6 – 6 – 1 = -1 \]

Check total charge. Sum the formal charges: plus one from nitrogen, zero from the double bonded oxygen, and minus one from each single bonded oxygen. The sum equals minus one, matching the ion charge. Because there are three equivalent resonance forms, the negative charge is delocalized over the three oxygens in the resonance-averaged structure.

Practical tips. Aim for Lewis structures where formal charges are close to zero and where negative formal charges reside on the more electronegative atoms. For isoelectronic choices, lower magnitudes of formal charge and separation of unlike charges is usually preferred.